Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y(3y - 10)}{8} \times \dfrac{9}{5(3y - 10)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ y(3y - 10) \times 9 } { 8 \times 5(3y - 10) } $ $ n = \dfrac{9y(3y - 10)}{40(3y - 10)} $ We can cancel the $3y - 10$ so long as $3y - 10 \neq 0$ Therefore $y \neq \dfrac{10}{3}$ $n = \dfrac{9y \cancel{(3y - 10})}{40 \cancel{(3y - 10)}} = \dfrac{9y}{40} $